Krull Dimension of Topological Subspace is Smaller

Definition

Let $X$ be a topological space.

Let $Y \subseteq X$ be a subspace.

Then:

$\map \dim Y \le \map \dim X$

where $\dim$ denotes the Krull dimension.

Proof

Let:

$A_0 \subsetneq A_1 \subsetneq \cdots \subsetneq A_n$

be a chain of closed irreducible sets of $Y$.

Let $\map \cl {A_i}$ be the closure of $A_i$ in $X$ for each $i = 0, \ldots, n$.

By Closure of Irreducible Subspace is Irreducible, each $\map \cl {A_i}$ is irreducible.

Furthermore:

$\map \cl {A_0} \subsetneq \map \cl {A_1} \subsetneq \cdots \subsetneq \map \cl {A_n}$

since, by Closure of Subset in Subspace:

$A_i = Y \cap \map \cl {A_i}$

Thus:

$n \le \map \dim X$

As $\map \dim Y$ is the spremum of such $n$, we have:

$\map \dim Y \le \map \dim X$

$\blacksquare$

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